Isosceles triangle 9

Given an isosceles triangle ABC where AB= AC. The perimeter is 64cm, and the altitude is 24cm. Find the area of the isosceles triangle.

Correct result:

A =  168 cm²

Solution:

$$\begin{gathered} p=64\text{ cm } \\ h=24\text{ cm } \\ p=2a+b \\ {a}^2=(b\mathrm{/}2{)}^2+h^2 \\ {a}^2=((p-2a)\mathrm{/}2{)}^2+h^2 \\ {a}^2=(p\mathrm{/}2-a{)}^2+h^2 \\ {a}^2=p^2\mathrm{/}4-pa+a^2+h^2 \\ pa=p^2\mathrm{/}4+h^2 \\ a=p\mathrm{/}4+h^2\mathrm{/}p=64\mathrm{/}4+24^2\mathrm{/}64=25\text{ cm } \\ b=p-2\cdot a=64-2\cdot 25=14\text{ cm } \\ {p}_1=2\cdot a+b=2\cdot 25+14=64\text{ cm } \\ {p}_1=p \\ A={\displaystyle \frac{b\cdot h}{2}}={\displaystyle \frac{14\cdot 24}{2}}=168{\text{cm}}^2 \end{gathered}$$

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Showing 2 comments:

Steve

No, your solution starts with h=25cm, but the question says that h=24cm.

So, looking at one-half of the isosceles triangle, x² + 24² = (32-x)²    solves to x=7cm,   which gives the area of the isosceles triangle as 7x24 = 168cm²

1 year ago  Like

Dr Math

thank you, we corrected 25 to 24 as altitude....

1 year ago  Like

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